Find the arithmetic progression and sum of its first 20 terms whose arithmetic mean and geometric mean between first and third term is 10 and 8 respectively
(a + a+2d)/2 = 10
√(a(a+2d)) = 8
a=4, d=6
So, the sequence is
4,10,16,...
S20 = 20/2 (2*4 + 19*6) = 1220
Im doing 8th grade statistics, i did 7th grade Geometry and Arithmetics before.
To find the arithmetic progression, we need to determine the first term (a) and the common difference (d).
Given that the arithmetic mean between the first and third terms is 10, we can use the formula for the arithmetic mean (mean = (first term + third term)/2) to express this relationship:
10 = (a + (a + 2d))/2
20 = 2a + 2d
10 = a + d [equation 1]
Given that the geometric mean between the first and third terms is 8, we can express this relationship using the formula for the geometric mean (mean = sqrt(first term * third term)):
8 = sqrt(a * (a + 2d))
64 = a * (a + 2d) [equation 2]
Now, let's solve the simultaneous equations 1 and 2. We can substitute the value of (a + d) from equation 1 into equation 2:
64 = a * (10)
64 = 10a
a = 64/10
a = 6.4
Substituting the value of a back into equation 1:
10 = 6.4 + d
d = 10 - 6.4
d = 3.6
Thus, the first term of the arithmetic progression is 6.4 and the common difference is 3.6.
To find the sum of the first 20 terms, we can use the formula for the sum of an arithmetic progression:
Sum = (n/2) * (2a + (n-1)d)
Substituting the values, where n = 20, a = 6.4, and d = 3.6:
Sum = (20/2) * (2 * 6.4 + (20 - 1) * 3.6)
Sum = 10 * (12.8 + 684)
Sum = 10 * 696.8
Sum = 6968
Therefore, the arithmetic progression is 6.4, 10, 13.6, 17.2, ..., and the sum of its first 20 terms is 6968.
To find the arithmetic progression and sum of its first 20 terms, we need to use the given arithmetic and geometric means.
Let's assume the first term of the arithmetic progression is "a", and the common difference is "d". Therefore, the second term is "a + d", and the third term is "a + 2d".
Given that the arithmetic mean between the first and third term is 10, we can write the equation:
(a + a + 2d) / 2 = 10
Simplifying this equation, we get:
2a + 2d = 20
a + d = 10
Now, let's calculate the geometric mean between the first and third term. The geometric mean is the square root of the product of these terms. So, we have:
√((a)(a + 2d)) = 8
√(a^2 + 2ad) = 8
a^2 + 2ad = 64
Now, we have a system of equations:
a + d = 10
a^2 + 2ad = 64
To solve this system of equations, we can substitute the value of "d" from the first equation into the second equation. It will allow us to find the value of "a".
Substituting a + d = 10 into a^2 + 2ad = 64, we get:
a^2 + 2a(10 - a) = 64
a^2 + 20a - 2a^2 = 64
-a^2 + 20a = 64
a^2 - 20a + 64 = 0
Now we can solve this quadratic equation to find the value of "a". Factoring or using the quadratic formula, we find:
(a - 4)(a - 16) = 0
a = 4 or a = 16
If a = 4, then the common difference is:
d = 10 - 4 = 6
If a = 16, then the common difference is:
d = 10 - 16 = -6
Now we have the possible arithmetic progressions:
1. First term (a) = 4, Common difference (d) = 6
2. First term (a) = 16, Common difference (d) = -6
To find the sum of the first 20 terms of an arithmetic progression, we can use the formula:
Sum = (n/2)[2a + (n-1)d]
For the first progression (a = 4, d = 6):
Sum = (20/2)[2(4) + (20-1)(6)]
Sum = 10[8 + 19(6)]
Sum = 10[8 + 114]
Sum = 10(122)
Sum = 1220
For the second progression (a = 16, d = -6):
Sum = (20/2)[2(16) + (20-1)(-6)]
Sum = 10[32 + 19(-6)]
Sum = 10[32 - 114]
Sum = 10(-82)
Sum = -820
Therefore, the arithmetic progression with a first term of 4 and a common difference of 6 has a sum of 1220 for the first 20 terms. The arithmetic progression with a first term of 16 and a common difference of -6 has a sum of -820 for the first 20 terms.